Probability

Hi, can someone help me with this probability problem?

A publisher is printing a 1200 page book. The machinery is good but not perfect, printing an unintended blotch on average once every 100 pages. A book will be rejected if there are 20 or more pages with a blotch. In a production run of 10,000, about how many books do you expect will be rejected?

I was told the answer is 102 (probability 0.0102 of a book's rejection) but i just can't seem to work it out. Your help is ver appreciated, and thank you in advance.

A publisher is printing a 1200 page book. The machinery is good but not perfect, printing an unintended blotch on average once every 100 pages. A book will be rejected if there are 20 or more pages with a blotch. In a production run of 10,000, about how many books do you expect will be rejected?

I was told the answer is 102 (probability 0.0102 of a book's rejection) but i just can't seem to work it out. Your help is ver appreciated, and thank you in advance.

The probability that a page is blotched is 0.01. There are 1200 pages in a
book, so the number of blotched pages has a binomial distribution:

P(n blotch)=B(n, 0.01, 1200)=1200!/(n!(1200-n)!) 0.01^n 0.99^(1200-n)

Now we could sum the appropriate number of terms of this,
(or rather use P(n>=20)=1-P(n<=19)=1-sum(n=0..19) P(n))
but I expect you are supposed to use the normal approximation
to the binomial at this point.

The normal approximation has a m= 0.01*1200=12 blotches, and a standard
deviation s=sqrt(1200*0.01*0.99)~=3.45.

In the normal approximation because it is continuous we translate 20 or
more as 19.5 or more, then we are asking for the probability of a normal
random variable of mean 12 and sd ~=3.45 exceeding 19.5.

(this approximation gives a probability of ~0.0148, rather than 0.102, so
I guess you are supposed to use the exact binomial probability.)